According to the Pythagorean theorem, BC = 4. From the metric relations in a right triangle, we find PB = AB * BC / AC = 12/5. The triangles ABC and PBQ are similar and their similarity ratio is 12/5/5 = 12/25. From this, PQ = 4 * 12/25 = 48/25 and BQ = 3 * 12/25 = 36/25. From this, the area of the triangle is (48/25*36/25)/2=864/625
by far and away the quickest way to do this is by recognising the similar triangles and that the areas in the ratios of corresponding sides squared. yes pythag gives you the base length and therefore the area is 6. The area of the left triangle is therefore 6*(9/25). Thus the area of the triangle on the right with base line as the hypotenuse is 12*(16/25) by subtraction. The partitioning of this triangle is a recursive treatment applied to what we just did so the triangle we want is 6*(16/25) * (9/25) = 864/625
864/625 or 1.3824 answer All three triangles are similar due to the right angles Since the triangle is 3-4-5 And AB = then BP is equivalent to the 4 ( of the 3-4-5) , hence 4/5 of 3 = 12/5 Since 12/5 is equivalent to 5 of the red-shaded triangle, then the other two bases are : 4/5 of 12/5 or 48/25 and 3/5 of 12/5 or 36/ 25 So, the red triangle is a 3-4-5 with the sides 12/5, 48/25, and 36/25 Scaled down by 12/25. That is if you divide all three sites of the red triangle by 12/25, you get a 3-4-5 triangle Hence, area of the red-shaded triangle is half, the base * height or 1/2 (48/25 * 36/25) = 1/2 * 1728/625 or 864/625
As CA = 5 and AB = 3, ∆ANC is a 3:4:5 Pythagorean triple right triangle and BC = 4. As ∠ABC = 90°, let ∠BCA = α and ∠CAB = β, where α and β are complementary angles that sum to 90°. As ∠CPB = 90°, ∠PBC = 90°-∠BCP = 90°-α = β. As ∠PBC = ∠CAB = β and ∠C = α is cpmmon, ∆CPB and ∆ABC are similar. As ∠CPB = 90°, ∠PBC = 90°-∠BCP = 90°-α = β. As ∠PBC = ∠CAB = β and ∠C = α is common, ∆CPB and ∆ABC are similar. As ∠BPA = 90°, ∠ABP = 90°-∠BPA = 90°-β = α. As ∠ABP = ∠BCA = α and ∠A = β is common, ∆BPA and ∆ABC are also similar. As ∠BQP = 90°, ∠QPB = 90°-∠PBQ = 90°-β = α. As ∠QPB = ∠BCA = α and ∠PBQ = ∠CAB = β, ∆BQP and ∆ABC are also similar. Triangle ∆BPA: BP/AB = BC/CA BP/3 = 4/5 BP = 3(4/5) = 12/5 Triangle ∆BQP: BQ/BP = AB/CA BQ/(12/5) = 3/5 BQ = (12/5)(3/5) = 36/25 QP/BP = BC/CA QP/(12/5) = 4/5 QP = (12/5)(4/5) = 48/25 [BQP] = bh/2 = BQ(QP)/2 [BQP] = (36/25)(48/25)/2 [BQP] = 18(48)/625 = 864/625 [BQP] = 1.3824 sq units
∆ ABC~∆CQO 3/5=PQ/CP So PQ=3a ; CP=5a AP=5-5a ∆ABC~∆APB (5-5a)/3=3/5 So a=16/25 So PQ=48/25 CP=80/25 So CQ=√(80/25)^2-(48/25)^2=64/25 BQ=4-64/25=36/25 So Red area=1/2(48/25)(36/25)=864/625.❤
ABC tiene lados 3/4/5→ BP=(3/5)*4=12/5→ Razón de semejanza entre BQP y ABC =s=(12/5)/5=12/25→ Razón entre sus áreas =s²=12²/25²=144/625→ Área BPQ=(3*4/2)*s²=6*144/625 =864/625 =1,3824 ud². Gracias y un saludo.
The area of the triangle inside us 864/625 units square. This might be another property od Pythagorean triples but I think-as was similar to yesterday's problem-that because of corresponding and alternating angles being congruent and complementary angles the progression of angles for all three triangles is alpha beta and 90°. Because all three right triangles are collinear!!! Someone review this comment please!!!
For jee, you need to solve as many questions as you can. Select any chapter and watch lecture video on it (available on youtube also) and then solve as many questions as you can on that chapter. If you are unable to solve then you can start practicing with solved examples. And also try to solve previous year questions of jee on that chapter. If you will do all this then no one can stop you to qualify in jee.
According to the Pythagorean theorem, BC = 4. From the metric relations in a right triangle, we find PB = AB * BC / AC = 12/5. The triangles ABC and PBQ are similar and their similarity ratio is 12/5/5 = 12/25. From this, PQ = 4 * 12/25 = 48/25 and BQ = 3 * 12/25 = 36/25. From this, the area of the triangle is (48/25*36/25)/2=864/625
by far and away the quickest way to do this is by recognising the similar triangles and that the areas in the ratios of corresponding sides squared. yes pythag gives you the base length and therefore the area is 6. The area of the left triangle is therefore 6*(9/25). Thus the area of the triangle on the right with base line as the hypotenuse is 12*(16/25) by subtraction. The partitioning of this triangle is a recursive treatment applied to what we just did so the triangle we want is 6*(16/25) * (9/25) = 864/625
864/625 or 1.3824 answer
All three triangles are similar due to the right angles
Since the triangle is 3-4-5
And AB = then BP is equivalent to the 4 ( of the 3-4-5) , hence 4/5 of 3 = 12/5
Since 12/5 is equivalent to 5 of the red-shaded triangle, then the other two bases are :
4/5 of 12/5 or 48/25
and 3/5 of 12/5 or 36/ 25
So, the red triangle is a 3-4-5 with the sides 12/5, 48/25, and 36/25 Scaled down by 12/25. That is if you divide all three
sites of the red triangle by 12/25, you get a 3-4-5 triangle
Hence, area of the red-shaded triangle is half, the base * height or 1/2 (48/25 * 36/25) = 1/2 * 1728/625 or 864/625
As CA = 5 and AB = 3, ∆ANC is a 3:4:5 Pythagorean triple right triangle and BC = 4.
As ∠ABC = 90°, let ∠BCA = α and ∠CAB = β, where α and β are complementary angles that sum to 90°.
As ∠CPB = 90°, ∠PBC = 90°-∠BCP = 90°-α = β. As ∠PBC = ∠CAB = β and ∠C = α is cpmmon, ∆CPB and ∆ABC are similar.
As ∠CPB = 90°, ∠PBC = 90°-∠BCP = 90°-α = β. As ∠PBC = ∠CAB = β and ∠C = α is common, ∆CPB and ∆ABC are similar.
As ∠BPA = 90°, ∠ABP = 90°-∠BPA = 90°-β = α. As ∠ABP = ∠BCA = α and ∠A = β is common, ∆BPA and ∆ABC are also similar.
As ∠BQP = 90°, ∠QPB = 90°-∠PBQ = 90°-β = α. As ∠QPB = ∠BCA = α and ∠PBQ = ∠CAB = β, ∆BQP and ∆ABC are also similar.
Triangle ∆BPA:
BP/AB = BC/CA
BP/3 = 4/5
BP = 3(4/5) = 12/5
Triangle ∆BQP:
BQ/BP = AB/CA
BQ/(12/5) = 3/5
BQ = (12/5)(3/5) = 36/25
QP/BP = BC/CA
QP/(12/5) = 4/5
QP = (12/5)(4/5) = 48/25
[BQP] = bh/2 = BQ(QP)/2
[BQP] = (36/25)(48/25)/2
[BQP] = 18(48)/625 = 864/625
[BQP] = 1.3824 sq units
BC=√[5²-3²]=4 BP=3*4/5=12/5 PQ=12/5*4/5=48/25 BQ=12/5*3/5=36/25
area of triangle BPQ = 48/25*36/25*1/2 = 864/625
BP = (AB x BC)/AC = 12/5; Area (BPQ)/Area(ABC) =( (12/5)^2)/(5^2),; Area(BPQ) = 864/25
∆ ABC~∆CQO
3/5=PQ/CP
So PQ=3a ; CP=5a
AP=5-5a
∆ABC~∆APB
(5-5a)/3=3/5
So a=16/25
So PQ=48/25
CP=80/25
So CQ=√(80/25)^2-(48/25)^2=64/25
BQ=4-64/25=36/25
So Red area=1/2(48/25)(36/25)=864/625.❤
ABC tiene lados 3/4/5→ BP=(3/5)*4=12/5→ Razón de semejanza entre BQP y ABC =s=(12/5)/5=12/25→ Razón entre sus áreas =s²=12²/25²=144/625→ Área BPQ=(3*4/2)*s²=6*144/625 =864/625 =1,3824 ud².
Gracias y un saludo.
3:4=b:h..3:4=h:(4-b)...h=48/25..b=36/25... Ayellow=bh/2=864/625
I have done it your way. Great.
@@user-xz7sv2dn9g👍
(3)^2=9 (5)^2=25 {9+25}=34 180°ABC/34=5.10 5.2^5 1.2^1 2^1 (ABC ➖ 2ABC+1).
Answer 1728/1250 or 864/625 or 1.3824
The area of the triangle inside us 864/625 units square. This might be another property od Pythagorean triples but I think-as was similar to yesterday's problem-that because of corresponding and alternating angles being congruent and complementary angles the progression of angles for all three triangles is alpha beta and 90°. Because all three right triangles are collinear!!! Someone review this comment please!!!
1.44 ×1.92=2.7648÷2=1.3824
864/625
Hi can you give some tips for class 11 mathematics (jee)?
It's tough to analyse as well as start questions so pls....
For jee, you need to solve as many questions as you can. Select any chapter and watch lecture video on it (available on youtube also) and then solve as many questions as you can on that chapter. If you are unable to solve then you can start practicing with solved examples. And also try to solve previous year questions of jee on that chapter. If you will do all this then no one can stop you to qualify in jee.
@@MathBooster thanks a lot
Could you share some questions on your channel as well?
問題の辺の値に対してもっと簡素な解答になるようにしろよ。