Question

In the adjoining figure, ABC is a right angled triangle at
A and AD is perpendicular to BC. If BC = 17 cm and
AC = 15 cm, find:
(1) the area of AABC.
(ii) the length of altitude AD.​

Answer 1

Step-by-step explanation:

AB^2+AC^2=BC^2( PYTHAGORAS PROPERTY)

AB^2=BC^2-AC^2

=17^2-15^2

=289-225

AB^2=64

AB=8

area of ABC=1/2×b×h

=1/2×AC×AB

=1/2×15×8

=15×4

=60cmsq

area of ABC=1/2×b×h

60=1/2×BC×AD

60=1/2×17×AD

AD=60×2/17

=120/17

=7.058

Answer 2

The given question is in the adjoining figure, ABC is a right-angled triangle at

A and AD is perpendicular to BC. If BC = 17 cm and

AC = 15 cm,

we have to find the

(1) the area of AABC.

(ii) the length of altitude AD.

This problem is solved by using a Pythagoras theorem

The theorem states that the In a triangle the sum of the square of two sides is equal to the square of the third side.

for the given figure, the expression based on Pythagoras theorem is

${ab}^{2} + {ac}^{2} = {bc}^{2}$

The value of ac and bc was known, then we have to find the value of AB, it was found as

${ab}^{2} = {bc}^{2} - {ac}^{2}$

substitute the values in the above formula

${ab}^{2} = {17}^{2} - {15}^{2} \\ = 289 - 225 = 64 \\ {ab}^{2} = 64 \\ ab = 8cm$

The formula to find the Area of the triangle ABC is

$\frac{1}{2} \times base \times height$

$\frac{1}{2} \times ac \: \times ab \\ \frac{1}{2} \times 15 \times 8 \\ = 15 \times \: 4 = 60cm \: sq$

Therefore, the answer for the 1 is 60 sq.

2. The length of the altitude AD.

The area of the triangle ABC is

$60 = \frac{1}{2} \times \: base \times height$

substitute the given values in the above expression, and we get

$60 = \frac{1}{2} \times bc \times ad$

$\frac{1}{2} \times 17 \times ad = 60 \\ad = \frac{60 \times 2}{17} \\ ad = \frac{120}{17}$

The value of the ad is 7.058.

Therefore, the length of the altitude is obtained as 7.058 cm

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