Can You Find the Area of this Semicircle? | Step-by-Step Explanation

[@martinmarc-olivier9077]

Very clear explanations. A faster way to reach the split k, 4k, is to notice that both triangles have the same height, therefore their bases must have the same ratio as their areas, that ratio being 28/7 = 4.

[@PreMath]

Thank you Martin So nice of you Keep it up dear👍

[@seema4380]

💮I have doubt in a question...help please . Question is: 4 isosceles triangles are inscribed in a semicicle circle{ triangle's base at semicircles base } all can be diff sizes)end to end joined and to both end of base of semicircles all triangles have same base angles (theta)..how to find theta

[@spafon7799]

Nice solution.

[@andreadanieli6192]

I love thinking in an alternative way! Instead of introducing the intersecting chords theorem I considered proportionality of green and purple triangles. n : h = h : m ---> h^2 = m * n ---> h^2 = 4 * m^2 ---> h = 2m Since m * h = 14 ---> m * 2m = 14 ---> m^2 = 7 ---> m = sqrt(7) Since n = 4m ---> n = 4 * sqrt(7) So we can obtain the diameter of the circle D = m + n = 5 * sqrt(7) Finally, area of the semi-circle is A = pi * D^2 / 8 = pi * 25 * 7 /8 = pi * 175/8 ; - )

[@sergiosereno1489]

Yes, my friend. I did it the same way.

[@johnbrennan3372]

Very nice method. Another way to get h in terms of k instead of intersecting chord theorem is to use the theorem Pythagorous on the three triangles.

[@srividhyamoorthy761]

Owesome sir , saw this late but enjoying

[@ОльгаСоломашенко-ь6ы]

It is clear that one more equation is needed to find the radius. We get from the similarity of right triangles n/h=h/m. mn= h^2. Multiply the first two equations, we get h^4=56*14. Further it is clear. The relationship between the height of a right triangle and the segments of the hypotenuse is studied in the school curriculum. The height is the average proportional between the projections of the legs on the hypotenuse. You probably have a different program from ours.

[@phungpham1725]

Thank you. I did it by using right triangle altitude theorem.

[@luigipirandello5919]

Beautiful solution. Thank you. Have a nice weekend, Sir.

[@basavarajchikkamath9070]

Very good problem, nice solution. Thank you sir. 👌🙏🙏

[@ayushshinde118]

Can we prove them similar and use area proportionality theorem of similar triangles

[@spafon7799]

I haven't checked the solution yet, but I would use similar triangles. The small triangle has small side a. Because the area of the larger triangle is 4x that of the small triangle then the base and height of the larger triangle are each 2x of those of the small triangle, hence the small side (height) of the larger triangle is 2a. Thus the small triangle has base a and height 2a. Thus its area is a^2=7 and thus a=sqrt(7). Since the triangles are similar, and the height/base of the larger triangle = the base/height of the small triangle, which is 1/2. Thus the base of the larger triangle is twice the height of that triangle 2a, and thus is 4a. The diameter of the semicircle is thus 5a, the radius is 2.5a, thus 2.5*sqrt(7). Hopefully. Now I'll check the solution.

[@kennethstevenson976]

Since both triangles have a common height and one area is four times the other then the green base is 4 times the purple base. The triangles are similar, producing the proportions h/k = 4k/h. Cross multiply you get h^2 = 4k^2 , simplify h=2k. Substituting green Area formula 28=1/2 x 4k x 2k ; 28=4k^2 ; 7 = k^2 and finally k= square root of seven. The radius is 5k/2 . The rest is the same without using the Intersecting Chord Theorem, which was a good thing to review. Math done correctly produces consistent results!

[@alessandropapi2120]

I used similarity of triangles instead of chord theorem. Good content

[@boanergesct5823]

The sum triangle of the two represented is a right triangle since it is inscribed in a semicircle (diameter 2R). Both triangles are similar since they have the three equal angles; their bases (“b” and “B”) add up to 2R and their lengths are in the same ratio as the areas (28/7=4), since they share the same height “h” → 4b=B → 2R=b+ B=b+4b=5b → 2R=5b → b=R2/5 → B=4b=R8/5. The similarity between both triangles allows us to write the following equality: b/h=h/B → (R2/5)/h=h/(8R/5) → h=R4/5. With the obtained values we can write: bh/2=7 → (R2/5)(R4/5)/2=7 → R²8/25=2x7=14 → R²=14x25/8 → Area of the semicircle = πR²/2 = π14x25/2x8 = π175/8 = 68.7223

[@Khalid8712]

Also we can do it by adding the A1and A2 which give us 35 and find h since it is 90° tringle.

[@sastipadasadhu2254]

You are genius sir

[@govindashit6524]

Another great solution. It's very helfful for all students. Thanks dear sir.

[@PreMath]

Many many thanks Govinda dear You are the best. Keep it up dear😀 Love and prayers from Arizona, USA!

[@ramaprasadghosh717]

Left part of Diameter/ Its Right part = 7/28 = 1/4 = D_L/ D_R say Hereby D_L = D/5 , D_R = 4D/5 Congruence triangle property gives H /D_R = D_L/ H or H^2 = D_L * D_R = 4 *D^2/ 5^2 or H = 2*D/5 Now H*D/2 = 7 + 28 = 35 so D^2/5 = 35 i.e D^2 = 175 Hereby area of semi circle π*D^2/8 = 175*π/8

[@gikasmith5511]

Nice. Although I think noticing that the entire triangle - green plus purple - is a right triangle (inscribed into a semicircle, inscribed angles property) would make the solution even easier. The two are therefore similar, so the height is the geometric mean of the projections - and voila.

[@highplayz]

I done it mentally by using similarity Theorem used : If in a right triangle Perpendicular drawn from 90° to hypotenuse then 2 triangles formed are similar to each other and to the whole triangle

[@242math]

very well done bro, thanks for sharing this area of a semicircle problem

[@brucecarter8296]

the 2 triangles are similar because the 2 chords make a 90 degree angle. the area of the larger is 4x that of the smaller, so the sides are 2x longer. if the short leg of the 7 triangle is x, then the short leg of the 28 triangle is 2x (also the long leg of the 7 triangle so the long leg of the 28 triangle is 4x) 1/2*x*2x=7 x=sqrt7 D=5x=5sqrt7

[@Mathematician6124]

It was awesome

[@beeruawana6662]

Very good question

[@flexeos]

when you have m and n =f(h) and 2r=a+b, you can also use directly r^2 = ((n-m)/2)^2 + h^2

[@SamsungJ-kk5nr]

Nice exercise.

[@imonkalyanbarua]

On applying the Chord's theorem you made an assumption that c=d which is why you wrote h^2=4k^2 but that is not the case as c=d will be a special case, we don't know that from the question, do we? Please correct me if I have made a mistake in my understanding. Thank you.

[@crabbyengineer274]

The initial problem statement should have included the assumption that both green and purple triangles are right triangles.

[@jimbobeire]

indeed, it then especially makes it easier to spot that they are similar triangles, and the area ration of 4 to 1 means their side ratios is 2 to 1, and also giving the sides as being root 7 and root 28. It's also then easy to work out that with the diameter of the semicircle is 5 times the short leg of the smaller triangle, the diameter is root 175, and the area of the semicircle is easy to obtain from that.

[@murdock5537]

see 0:43...

[@KhushbuGupta-pl6fj]

Thank you so much sir . Your videos help me a lot . I spend my 1 hour daily on your you tube channel

[@PreMath]

Glad to hear that Khushbu dear. Thanks for awesome feedback. You are the best. Keep it up dear😀 Love and prayers from Arizona, USA!

[@learnenglishwithsanafarhan]

Very nice content 💯🍀

[@NASIR58able]

Well done,

[@ramanivenkata3161]

Very well explained.👌👌

[@PreMath]

Thanks a lot 😊 So nice of you Ramani Keep it up dear👍

[@tahasami597]

Thank for premath

[@frankjohn1408]

Question; how can you assume that there is a right angle between the two triangles? Where they meet on the circle is a right angle but the other cannot be assumed. What am missing?

[@TurquoizeGoldscraper]

I used Thales' Theorem to show that the big triangle was a right triangle and that all three triangles where congruent, and then used their ratios to get h = 2k.

[@misterenter-iz7rz]

2a^2=14, 8a^2=56, a^2=7, 5a is the diameter, 5a/2 is the radius, therefore the answer is (1/2)(5a/2)^2pi=25/8pia^2=175/8 pi=68.7 approximately.

[@sanjeevipandurangan9836]

The problem can also be solved using cosine formula. The angle subtended by the diameter is 90 degrees. So cos 90 is equal to zero.

[@wrigjo101]

How do ne know that the triangles are right angles from the information provided in the question?

[@oseznos]

Why this angle is a 90 degrees angle?

[@PreMath]

Dear Jose, Thales' theorem: Inscribed angles subtended by a diameter are always 90 degrees! Thanks for asking. Keep it up dear👍

[@oseznos]

@@PreMath First of all, thanks for your quick answer. And a very great apologice for my bad english. But I don't agree with your explanation.. I know Thales Therem, but here Thales DO NOT SAY ANYTHING ABOUT THE DRAW. In maths, you know better than me, It's not possible to suppose anithing about the draws. Tha "h" segment is perpemdicular to the other side because YOU DRAW IT PERPENDICULAR. But the problem DO NOT SAY US anything about this angle. Obviously if you do not suppose this segment is perpenducular, the problem has not solution but I think you must no suppose things to solve any problem. Thanks for all the problmes you give to us avery week.

[@holyshit922]

This picture reminds me construction of segment with length of the side equal to te geometric mean of lengths of this two sides on which diameter is divided To proove that this is correct way of constructing geometric mean we can use similar triangles We find similar triangles with angle-angle-angle criterion and set proper proportion

[@tasdikabilal6170]

I like this type of question it is very interesting ....😊

[@PreMath]

Glad you liked it Keep it up Bilal dear👍

[@charlesbromberick4247]

Profe: Here´s a problem you may find interesting and somewhat challenging: Three circles of radii 1, 2 and 3 are externally tangent to one another. A fourth circle is inscribed in the space between the other three. What is the radius os this fourth circle? (Answer: 6/23)

[@holyshit922]

Draw radius from center to vertex outside diameter Let x be the base of purple triangle Calculate h in terms of x from formula of area of this purple right trangle From Pythagorean theorem in right triangle with sides h, R-x ,R we calculate R in terms of x From formula of area of right triangle coloured on green we calculate R once we have R , it is easy to calculate area of semicircle

[@leandrobastos329]

Hi, i really like your video. But I have a Question, could it be the radius is just five ? Because if you take : (7+28)2 = 70 the complete rectangle (think out of the circle) rectangle area: Big side x small side = base x height = 70 70 (1, 2, 5, 7, 10, 14, 35, 70) the base (big side) = 1x + 4x (because 7+28) try 7 * 10 =70 so x equal 2 4x equal 8 2*7/2 =7 8*7/2=28 2+8 = 10 (diameter) 10/2 =5 radius area = π5^2/2 12,5π

[@bienvenidos9360]

No, the radius can't be 5. If the radius were 5, then the maximum height will be 5, but there is a height of 7 which is above that amount.

[@santiagoarosam430]

28/7=4 ⇒ 2R=K+4K=5K ⇒ R=5K/2; KH=2x7=14 ⇒ H=14/K. If the lower end of H is the point "P", the power of "P" with respect to the circumference allows us to say that: H²=Kx4K=4K² ⇒ 14²/K²=4K² ⇒ K²K²=196/4=49 ⇒ K=√ 7 ⇒ R=5K/2=5√7/2

[@Roberto74B]

i have used a more simply logic ... when you have found the relation K->2K->4K you could imaging a double square of 56 and obtain the 2k doing the sqrt(28) ...

[@theophonchana5025]

#intersectingChordsTheorem #intersectingchords #chordstheorem #Chords

[@MrPaulc222]

((8/5)r)/h = h /((2/5)r ((16/25)r^2 = h^2 h = (4/5)r (4/5)r * (2/5)r = 14 (8/25)r^2 = 14 r^2 = (14*25)/8 = 350/8 = 175/4. r = (sqrt(175))/2 = (5*sqrt(7))/2 As r = sqrt(175)/2, r^2 = 175/4. Full circle = (175/4)pi Semicircle = (175/8)pi. Area approx 68.72 un^2. I see yours is 68.69, but I used a slightly more accurate value for pi.

[@NaRahul]

Im happy that im able to solve this problem and got 68.75.. answer

[@himo3485]

purple triangle ∞ green triangle 7 : 28 = 1 : 4 = 1² : 2² purple triangle = 1 = a green triangle = 2 = 2a a , 2a ,4a a+4a=5a purple triangle + green triangle = 7 +28 = 5a･2a･1/2 5a² = 35 a²=7 a>0 , a=√7　 radius : 5a/2 = 5√7/2 Area of the semi-circle : 5√7/2 ･ 5√7/2 ･ π ･ 1/2 = 175π/8

[@andirijal9033]

sir please help me, if x + y + z = 2 and xy + xz + yz = 1, what is maximum value (x - z ) ?

[@nicholascage3699]

U got me interested in math ...I am adult but love ur explanation

[@PreMath]

My dear friend Nicholas, we are all lifelong learners. Thanks for awesome feedback. You are the best. Keep it up dear😀 Love and prayers from Arizona, USA!

[@susennath6035]

Nice

[@freeguy77]

After you calculate the answer of 68.69 sq units, the white remainder is then calculated as: 68.69 - 7 - 28 = 33.69.

[@amzingcloud]

🤔🤔🤔 How you can do h^2 = 4k^2, as given circle ab = cd but a is not equal to b or c is not equal to d, so with this logic you can not do a^2 because a is not equal to b. So how you can do h^2....please explain...

[@davidfromstow]

Excellent!

[@PreMath]

Thank you David! Cheers! Keep it up dear👍

[@vaghelabhikhusing6553]

🤩🤩🤩🤩🤩 good information sir 🙏🙏🙏🙏

[@PreMath]

So nice of you Vaghela Keep it up dear👍

[@dreael]

You also can use the Thales's theorem, i.e. the bottom angle is also 90 degrees. With the angles you can show that the green and purple triagle are similar, even the resulting triangle of adding both triangles is similar. So because of that, h must be the geometric mean of m and n.

[@theoyanto]

What can I say, superlatives fail me, but I'll try... Exceptional example, above and beyond the call of duty, you exceeded even your own high standard, what a fantastic example, there was no way I could have completed this without your video. Thanks again 👍🏻

[@phanimaheswara7492]

Thank you sir

[@PreMath]

Most welcome Phani dear Keep it up dear👍

[@alvarosarmientogomez4996]

Míster Teacher. ...How about putting it all ( m, n, h ) based on the radius R ?... First.- In triangles of equal height, their areas are proportional to their respective bases. And let"s put "2R - m" instead of "n". So : m/7 = ( 2R - m )/28 28 m = 14 R - 7 m m = 14 R/ 35 = 2R/5 And : 2 R - m = 2 R - 2R/5 = 8R/5 Second. - Application of the "string theorem" to calculate "h" as a function of "R" : 2R/5 . 8R/5 = h2 And solving : h = 4R/5 Third and finally : In the triangle on the left ( for instance ) we have that m h / 2 = 7 Then : ( 2R/ 5 ). ( 4R/ 5 ) / 2 = 7 4 R2 / 25 = 7 4 R2 = 175 R2 = 175/4 Then, the area of the semicircle will be : ( "Pi" 175/4 ) . ( 180/360 ) = "Pi" 175 / 8 = 68.69 u2 Greetings from Bilbao ( Spain ), and stay blessed.

[@seema4380]

💮I have doubt in a question, how can I contact him... any email or something...help please 🙏 Question is: 4 isosceles triangles are inscribed in a semicicle circle{ triangle's base at semicircles base } all can be diff sizes)end to end joined and to both end of base of semicircles all triangles have same base angles (theta)..how to find theta

[@dukemmerife9310]

68.75 (pi = 22/7)

[@adgf1x]

At of semi circle=25pai/4 sq when pai=22/7

[@hrishikeshkashyap3715]

Let, hypotenuse of the purple triangle = a Hypotenuse of the green triangle = b Let, the hypotenuse of the largest triangle = c + d, where, "c" is the base of the green triangle & "d" is the base of the purple triangle. Let, the height of the green triangle and purple triangle = h. Thus, Area of green triangle = ch/2, 28 = ch/2.... So, c = 56/h. Also, Area of purple triangle = dh/2, 7 = dh/2.... So, d = 14/h In green triangle, using Pythagoras theorem, b^2 = c^2 + h^2 b^2 = (56/h)^2 + h^2 b = [(56/h)^2 + h^2]^(1/2) Now, Area of the largest triangle = ab/2, 35 = ab/2, 70 = ab.... So, a = 70/b. But, b = [(56/h)^2 + h^2]^(1/2). Thus, a = 70 ÷ [(56/h)^2 + h^2]^(1/2). Finally, in purple triangle, using Pythagoras theorem, a^2 = d^2 + h^2, where, d = 14/h & a = 70 ÷ [(56/h)^2 + h^2]^(1/2). This is how we can find "h" (altitude of the largest triangle *from its hypotenuse*).

[@theophonchana5025]

#Radius

[@mdavis9311]

Whoa, whoa-- chord theorem shows ab as shorter than diameter, cannot equate to hh. What tells you cd is on the diameter?

[@geraldillo]

Nice!

[@PreMath]

Thank you! Cheers!😀 Love and prayers from Arizona, USA!

[@r.a.6459]

Instead of using k, just use m from the fact that m/n = ¼ which leads to n = 4m and diameter m + 4m = 5m.

[@theophonchana5025]

Radius = (5 × square root of 7) ÷ 2

[@theophonchana5025]

D = 5 × square root of 7

[@vidyadharjoshi5714]

why keep sqrt 7 all the way ? sqrt 7 = 2.645 times 5 = 13.23, so r = 6.614; r sq = 43.75, times pi = 137.44 ( area of circle ). half = 68.69

[@cedarriverbill1719]

Sure took the long way around on that one.

[@theophonchana5025]

r = (5 × square root of 7) ÷ 2

[@christopherellis2663]

Hemi- demi--semi-quaver Hemisphere, Demigod, Semicircle

[@viniciusgarcez5401]

The image don't give you any clue that The common side of both triangles is the height relelated to The diameter of the semi-circle (forming the 90 degrees angle to it), If it does and knwoking that every single triangle drawing inside the semi-circle has 90 degrees angle to the opppsite side of the diameter it would be easier to find the radius and the area of the semi-circle

[@philipkudrna5643]

Before watching: from the similar triangles and some relationships and algebra, I derived that the base of the smaller triangle is sqrt(7), base of the larger triangle is 4*sqrt(7), the height of both triangles is 2*sqrt(7). (Not needed, but for checking purposes.) it follows that the sum of the bases (which is the diameter of the semicircle) is 5*sqrt(7). Thus the area of the semicircle is 175/8*Pi (or 68,72) square units. After watching: I didn‘t use the intersecting chords theorem but the similar triangles, but I still pretty much nailed it! 😀

[@mustafizrahman2822]

Failed to solve it. I have tried many methods but failed. Even also by orthocenter but failed.😞😢 However, it is a fine question.👍

[@PreMath]

No worries! Keep it up dear👍

[@theophonchana5025]

A = 28

[@alvarosarmientogomez4996]

Go to bed. ...

[@harikatragadda]

Purple triangle(base= a), and green triangle (base=b) are similar. a/h=h/b h²=ab We also have ½ah=7 and ½bh=28 Adding them, ½h(a+b)= hR=35 R=35/h Multiplying them, ¼h²ab=7*28=¼h⁴ Area of semicircle=½πR² = ½π35²/h²=(175/8)π

[@PreMath]

Excellent Hari! Keep it up dear👍

[@vsevolodtokarev]

Like your simpler method.

[@theophonchana5025]

A = 7

[@alvarosarmientogomez4996]

Go to bed. ..

[@Teamstudy4595]

Easiest question!!

[@calspace]

Too complicated. The green and purple triangles are similar. In a similar triangle you have sides a, b, and c and A, B, and C, with A=xa, B= xb, and C=xc. The area of the smaller triangle is 1/2 * ab and the area of the larger triangle is 1/2 * AB = 1/2 * x² * ab. Here 1/2 * ab = 7 and 1/2 * x² * ab = 28 or x² = 4, so x = 2. So n = 2h and h = 2m, or n = 4m So the area of both triangles, 7 + 28 = 35, is equal to 1/2 * (4m + m) * 2m = 5m². 5m² = 35 m² = 7 m = sqrt(7) 5m = 5sqrt(7)

[@Teamstudy4595]

Same question with different figures I Solved yesterday Night before Sleeping from another channel.....

[@Teamstudy4595]

Solved in few seconds

[@Teamstudy4595]

Ans : Area of semi circle = 175/8 Pie

[@PreMath]

Excellent Jayant Keep it up dear👍

[@Teamstudy4595]

@@PreMath Thank you SIR

[@DhirajKumar-sg3wn]

Method,of, explanquation,is,uniqe,sir

[@Tjen79]

If you use the theorem of thales you can see the lower angle of both triangles together is 90 degrees. Therefor both triangles are congruent. The area is 4x bigger so with the magnificationfactor of an area in mind you can conclude that the hight is 2x bigger then the purple base. This implies the the green base is 4x the base of the purple one. No need for calculations there just pure simple logics 😉 (edit its

[@ggeshundra]

Why is angle between h and m is 90°

[@PreMath]

Hello dear, Thales' theorem: Inscribed angles subtended by a diameter are always 90 degrees! Thanks for asking. Keep it up dear👍

[@ggeshundra]

@@PreMath wait, I don't understand how the angle between h and m is inscribed, as well how is it suspended by the diameter. Thank you in advance

[@vsevolodtokarev]

@@PreMath No, it's not the inscribed angle; it's the angle between the height and the diameter. It must be given in order for the problem to make sense.

[@theophonchana5025]

Area = 175 pi ÷ 8

[@PreMath]

Excellent Theo! Keep it up dear👍

[@nivan231]

68.75 is the right answer.. we shouldn't put the value of π as 3.14, instead it should be 22/7.. 175 is easily divisible by 7.. tx

[@freeguy77]

Saying it another way, carry the PI value out to more decimal places than only 2 decimals: 3.14159265. That is more accuracy than is usually required. 1/7th is not the same number, and will create errors after the first decimal place: 1/7 = 0.142857142857... so the difference in the decimal side of the two is: 0.00126449. The closest answer, to 3 decimal places is: 175 * 3.14159265 / 8 = 68.722. So you are off by almost 0.03 sq units!

[@williamwingo4740]

As usual, here's a slightly different approach. Someone else already observed that the purple and green triangles have the same height, so the bases are proportional to the areas: that is, k for the purple and and 28k/7 or 4k for the green. Considering the purple triangle, we have kh/2 = 7, so kh = 14, so h = 14/k. Invoking Pythagoras, the square of the hypotenuse of the purple triangle (call it y) is y^2 = k^2 + (14/k)^2. Similarly, the square of the hypotenuse of the green triangle (call it z) is z^2 = (4k)^2 + (14/k)^2 = 16k^2 + (14/k)^2. There's no need to take square roots to get y and z because we're going to be squaring them again almost immediately anyway. The diameter of the semicircle is the hypotenuse of the big triangle and is equal to 5 k. Again calling up Pythagoras, we have (5k)^2 = y^2 + z^2 = k^2 + (14/k)^2 + 16k^2 + (14/k)^2; simplifying and multiplying out, this becomes 25k^2 = k^2 + 196/k^2 + 16 l^2 + 196/k^2 = 17k^2 + 2(196/k^2). Simplifying again, we get 4k^2 = 196/k^2; multiplying through by k^2: 4k^4 = 196; k^4 = 196/4 = 49; and finally, k = 4th root of 49 = sqrt(sqrt(49)) = sqrt(7). From this point we proceed conventionally: area of semicircle = (pi(5k/2)^2)/2 = (pi)((25/4)k^2)/2 = (pi)(25/8)k^2 = (pi)(25)(7)/8 = 175 pi/8. Carpe Diem. 🤠