Question

in the adjoing figure abcd is a square of side 10 cm, p is the midpoint of the bc and apq is a right triangle, right angled at p. area of apq is :
a) 30.8 cm²
b) 35.79 cm²
c) 40 cm²
d) 50 cm²

Answer 1

Given :-----

• ABCD is a Square with side 10cm.
• P is mid point of BC .
• APQ is a right angle ∆ .

To Find :------

• Area of ∆APQ .

Formula used :-----

• Pythagoras theoram .
• Area of ∆ = 1/2 × Base × Height .
• Area of Square = side × side ...

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we can solve this by 2 methods :-----

→ 1) We can Find area of ∆ ABP, ∆ADQ, ∆QCP , and we will subtract from area of Square . we will get area of ∆APQ.

→ 2) we can find length of AP and PQ , by pythagoras theoram and Than find Area of ∆APQ...

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$\Large\bold\star\underline{\underline\textbf{Solution(1)}}$

it is given that, P is mid - point of BC .

so,

→ BP = PC = 5cm.

and,

→ DQ = DC - QC = 10-4 = 6cm.

$\pink{\large\boxed{\bold{Area of \triangle = 1/2 \times Base \times perpendicular}}}$

so,

→ Ar.[∆ABP] = 1/2 × 10×5 = 25cm²

→ Ar.[ADQ] = 1/2 × 10×6 = 30cm²

→ Ar.[QCP] = 1/2 × 4 × 5 = 10cm²

Now,

$\green{\large\boxed{\bold{square \: area = {side}^{2} }}}$

so,

→ Ar[ABCD] = 10×10 = 100cm² .

Hence, ∆APQ Area = 100-(25+30+10) = 35cm² ...

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$\Large\bold\star\underline{\underline\textbf{Solution(2)}}$

Lets try to Find length of AP and PQ, by pythagoras theoram ,,,

→ AP² = AB² + BP²

→ AP = √(10²+5²) = 5√5cm²

similarly,

→ PQ = √4²+5² = √41 cm²

so,

Ar[∆APQ] = 1/2 × 5√5 × √41 = 11.18 × 3.12 = 35cm² [ Approx] ..

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Answer 2

$\huge\bold\green{AnSwEr:}$

It is given that, P is mid - point of BC .

so,

→ BP = PC = 5cm.

→ DQ = DC - QC = 10-4 = 6cm.

By using formula , AREA OF TRIANGLE

$\large \sf \green{ \frac{1}{2} \times base \times height}$

so,

→ Area [∆ABP] = 1/2 × 10×5 = 25cm²

→ Area [ADQ] = 1/2 × 10×6 = 30cm²

→ Area [QCP] = 1/2 × 4 × 5 = 10cm²

Now,

$\sf\green{Square \: area = {side}^{2} }$

so, Area of [ABCD] = 10×10 = 100cm² .

Hence,

Area of ∆APQ = 100-(25+30+10) = 35cm²