Question

7. ΔABC is right angled at A (Fig 11.25). AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, Find the area of ΔABC. Also find the length of AD.
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Answer 1

Answer:

30 cmsq is the right answer

Answer 2

Given:-

• AB = 5cm
• BC = 13cm
• AC = 12cm
• AD = ?

To Find:-

• Area of the figure.
• Length of AD.

Solution:-

The given figure is a triangle with measures of each side is different.So, it is a scalene triangle.

We know that to find the area of a scalene triangle we use the Heron's Formula.

The Heron's Formula is :-

• $\footnotesize\underline{\sf{ \longmapsto \: \sqrt{s( s- a)(s -b )(s - c)} }}$

Where,

• s = a+b+c/2
• a = first side
• b = second side
• c = third side

On calculating the s we get:-

• $\footnotesize{\bf{ \frac{12cm + 5cm + 13cm}{2} }}$
• $\footnotesize{\bf{ \frac{30cm}{2} }}$
• $\footnotesize\underline{\bf{15cm}}$

So, now we will find the area of the figure:-

• $\footnotesize\underline{\bf{ \sqrt{ \{15(15 - 13)(15 - 12)(15 - 5) \}} {cm}^{2} }}$
• $\footnotesize\underline{\bf{ \sqrt{ \{3 \times 5(2)(3)(2 \times 5) \}} {cm}^{2} }}$
• $\footnotesize\underline{\bf{ \longmapsto3 \times 5 \times 2}}$
• $\large \green{\underline{ \blue {\boxed{\bf{ \maltese \: \: \: \: \red{30cm^{2} }}}}}}$

Now we got the area of the figure.

Now we will use the formula that is acceptable to all the triangle.

• ½×b×h

Where,

• b = base = 13cm
• h = height = AD
• Area = 30cm²

On substituting the values we get:-

• $\large\underline{\bf{ \frac{1}{2} \times 13cm \times h = 30 {cm}^{2} }}$
• $\large\underline{\bf{h = \frac{30 {cm}^{2} \times 2}{13cm} }}$
• $\large\underline{\bf{h = \frac{60cm}{13} }}$
• $\large\underline{\bf{h = 4.61cm}}$