Question

triangle ABC is a right angled triangle eith angle B = 90° . triangles PAB , QAC ,RBC , are constructed on sides AB,AC and BC respectively . prove that (area of triangle PAB) +(area of triangle RBC )=(area of triangle QAC​

Answer 1

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( SORRY FOR LEAVING THE QUESTION LIKE THAT ...)

★ GIVEN ★

• TRIANGLE ABC IS RIGHT ANGLE TRIANGLE

RIGHT ANGLED AT B

• triangles PAB , QAC ,RBC , are constructed on sides AB,AC and BC respectively

• TRAINGLE PAB , QAC , RBC ARE EQUALITERAL TRIANGLES

★ FIGURE ★

• REFER TO THE ATTACHMENT

★ TO PROVE ★

AREA OF TRIANGLE ( PAB + RBC )

= AREA OF TRIANGLE QAC

★ PROOF ★

ACCORDING TO THE QUESTIONS

→TRIANGLE ( PAB , RBC , QAC ) ARE EQUALITERAL TRIANGLES

(GIVEN )

→HENCE TRIANGLE ( PAC ≈ RBC ≈ QAC )

( BY AA SIMILARITY AS EACH ANGLE 60°)

THEREFORE ,

→ IN TRIANGLE ( RBC AND QAC )

$\frac{area(rbc)}{area(qac)} = \frac{ {bc}^{2} }{ {ac}^{2} } \: \: \: \: \: \: be \: equation \: 1$

( by area ratio THEOREAM )

→ IN TRIANGLE ( PAB AND QAC )

$\frac{area \: (pab)}{area \:(qac)} = \frac{ {ab}^{2} }{ {ac}^{2} } \: \: \: \: \: \: be \: equation \: 2$

( by area ratio THEOREAM )

→ ADD EQUATION 1 AND 2

$\frac{area(rbc)}{area(qac)} \: + \frac{area \: (pab)}{area \: (qac)} = \frac{ {bc}^{2} }{ {ac}^{2} } + \frac{ {ab}^{2} }{ {ac}^{2} }$

→ NOW BY ADDING FURTHER WE GET...

$\frac{area(rbc) + area(pab)}{area(qac)} = \frac{ {ab}^{2} + {bc}^{2} }{ {ac}^{2} }$

→ now AB² + BC ² = AC²

( BY PHYTOGORAS THEOREAM )

$\frac{area(rbc) + area(pab)}{area(qac)} = \frac{ {ac}^{2} }{ {ac}^{2} }$

$\frac{area(rbc) + area(pab)}{area(qac)} = 1$

HENCE ....

AREA ( RBC ) + AREA ( PAB ) = AREA ( QAC )

HENCE PROVED......

HOPE IT HELPS THANKS FOR YOUR QUESTION...

★ KEEP SMILING ☺️✌️ ★