Question

The base and hypotenuse of a right triangle is 5cm and 13 cm. find its area
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Answer 1

$\huge{\underline{\mathtt{\blue{Question}}}}$

The base and hypotenuse of a right triangle is 5cm and 13cm long. find its area

$\huge{\underline{\mathtt{\blue{Answer}}}}$

(refer the attachment )

GIVEN

• $\sf{PQR \:is \:right \:angle \:Triangle}$
• $\sf{Base (QR) = 5cm}$
• $\sf{Hypotenuse (PR) = 13cm}$

TO FIND

• $\sf{Hight (PQ) = ?¿}$
• $\sf{Find \:the \:area \:of \:∆PQR}$

We know the formula ;

$\sf{By \:phythagorus \:theorm}$

${\green{\overline{\green{\underline{\blue{\boxed{\orange{\mathtt{(Hypotenuse)² = (base)² × (height)²}}}}}}}}}$

$\sf{⇝ (PR)² = (QR)² + (PQ)²}$

$\sf{⇝ (13)² = (5)² + (PQ)²}$

$\sf{⇝ 169 = 25 + (PQ)²}$

$\sf{⇝169 - 25 = (PQ)²}$

$\sf{⇝169 - 25 = (PQ)²}$

$\sf{⇝144 = (PQ)²}$

$\sf{⇝\sqrt{144} = (PQ)}$

$\sf{⇝12 = (PQ}$

• PQ = 12cm

Now let's find Area of the triangle ;

We know the formula ;

${\green{\overline{\green{\underline{\blue{\boxed{\orange{\mathtt{Area = \frac{1}{2} × base × height}}}}}}}}}$

$\sf{⇝A = \frac{1}{2} × 5 × 12}$

$\sf{⇝A = \frac{1}{\cancel{2}^{1}} × 5 × \cancel{12}^{6}}$

$\sf{⇝A = 5 × 6}$

$\sf{⇝A = 30cm²}$

$\sf\red{Here \:we \:get \:your \:answer :))}$

Answer 2

$\huge \fbox \red{❥Question}$

The base and hypotenuse of a right triangle is 5cm and 13cm long. Find its area.

$\huge \fbox \green{❥Answer}$

(refer the attachment )

GIVEN

$\sf{PQR \:is \:right \:angle \:Triangle}$

$\sf{Base (QR) = 5cm}$

$\sf{Hypotenuse (PR) = 13cm}$

$\sf{Hight (PQ) = ?¿}$

$\sf{Find \:the \:area \:of \:∆PQR}$

We know the formula ;

$\sf{By \:phythagorus \:theorm}$

${\green{\overline{\green{\underline{\blue{\boxed{\orange{\mathtt{(Hypotenuse)² = (base)² × (height)²}}}}}}}}}$

$\sf{⇝ (PR)² = (QR)² + (PQ)²}$

$\sf{⇝ (13)² = (5)² + (PQ)²}$

$\sf{⇝ 169 = 25 + (PQ)²}$

$\sf{⇝169 - 25 = (PQ)²}$

$\sf{⇝169 - 25 = (PQ)²}$

$\sf{⇝144 = (PQ)²}$

$\sf{⇝\sqrt{144} = (PQ)}$

=(PQ)

$\sf{⇝12 = (PQ}$

PQ = 12cm

Now let's find Area of the triangle ;

We know the formula ;

${\green{\overline{\green{\underline{\blue{\boxed{\orange{\mathtt{Area = \frac{1}{2} × base × height}}}}}}}}}$

$\sf{⇝A = \frac{1}{2} × 5 × 12}$

$\sf{⇝A = \frac{1}{\cancel{2}^{1}} × 5 × \cancel{12}^{6}}$

$\sf{⇝A = 5 × 6}$

$\sf{⇝A = 30cm²}$

$\huge \mathtt \purple{@VioletMoon}$