Question

the base of a right angled triangle is 5 cm and hypothenuse is 13 cm . find its area​

Answer 1

Answer:

Given :-

• The base of a right angled triangle is 5 cm and hypotenuse is 13 cm.

To Find :-

• What is the area of a right angled triangle.

Formula Used :-

$\clubsuit$ Area Formula :

$\mapsto \sf\boxed{\bold{\pink{Area_{(Triangle)} =\: \dfrac{1}{2} \times Base \times Height}}}$

Solution :-

First, we have to find the height (AB) :

Given :

• Base (BC) = 5 cm
• Hypotenuse (AC) = 13 cm

According to the Pythagoras Theorem Formula we have to find the value of AB :

$\leadsto \bf{(AB)^2 =\: (AC)^2 - (BC)^2}$

$\leadsto \sf (AB)^2 =\: (13)^2 - (5)^2$

$\leadsto \sf (AB)^2 =\: (13 \times 13) - (5 \times 5)$

$\leadsto \sf (AB)^2 =\: 169 - 25$

$\leadsto \sf (AB)^2 =\: 144$

$\leadsto \sf AB =\: \sqrt{144}$

$\leadsto \sf\bold{\purple{AB =\: 12}}$

Now, we have to find the area of a right angled triangle :

Given :

• Base (BC) = 5 cm
• Height (AB) = 12 cm

According to the question by using the formula we get,

$\longrightarrow \sf Area_{(Triangle)} =\: \dfrac{1}{2} \times 5 \times 12$

$\longrightarrow \sf Area_{(Triangle)} =\: \dfrac{1}{\cancel{2}} \times {\cancel{60}}$

$\longrightarrow \sf\bold{\red{Area_{(Triangle)} =\: 30\: cm^2}}$

${\small{\bold{\underline{\therefore\: The\: area\: of\: a\: right\: angled\: triangle\: is\: 30\: cm^2\: .}}}}$

Answer 2

Answer:

$\bold{30cm {}^{2} }$

Step-by-step explanation:

$\large \qquad\dag \frak{Given:}$

$\sf \: Base(YZ)=5 \: \: cm$

$\sf \: Hypotenuse \: (XZ)=13 cm$

$\sf \: Let's \: find \: the \: length \: of \: XY \: first$

$\:$

$\large \qquad\dag \frak{Solution}$

$\sf \: By \: Pythagoras \: Theorem$

$\sf(The\: length \: of \: hypotenuse \: is \: equal \: to \: the\: Sum \: of \: other \: 2 \: sides)$

$\tt \qquad \: →XY {}^{2} +YZ {}^{2} =XZ {}^{2}$

$\tt \qquad \: →XY {}^{2} =XZ {}^{2} - YZ {}^{2}$

$\tt \qquad \: →XY {}^{2} = 169 - 25$

$\tt \qquad \: →XY {}^{2} = \sqrt{144} cm$

$\tt \qquad \: →XY {}^{2} = 12 \: cm$

$\:$

$\sf \: Now, As \: we \: know \: that$

$\bold \red{Area \: of \: triangle} \bold{= \frac{1}{2} \times Base \times height}$

$\sf {Area \: of \: triangle} = \frac{1}{2} \times 5 \times 12$

$\bold{Area \: of \: triangle} = \bold \red{30 {cm}^{2} }$

$\:$

$\bold \red{Celestial} \bold{Centrix}$