Question

If PQR is a right angled traingle at P, PS perpendicular QR. If PQ=5cm and PR=12cm,find the area of ∆PQR.also find the length of PS

Answer 1

ar. of triangle = 1/2 base*height
area of PQR=1/2*5*12
area=30cm^2
QR=13 by Pythagoras theorem
30 = 1/2 *PS * 13
PS= 60/13

Answer 2

GIVEN:

A triangle Q P  R right angled at P

Length P Q=5cm (height)

Length P R=12cm (foot)

P S is perpendicular to Q R

TO FIND:

The area of triangle Q P R, also find the length of P S

SOLUTION:

Area of triangle

$=\frac{1}{2}foot*height\\=\frac{1}{2} 12*5\\=30cm^{2}$

Now for length P R, we will use the Pythagoras, which will act as the new foot for the triangle Q P R

According to Pythagoras

$hypotenuse^{2} =foot^{2} +perpendicular^{2} \\QR^{2} =5^{2}+ 13^{2} \\QR=\sqrt{25+144} \\QR=\sqrt{169}\\QR=13cm$

Since now we got Q R=13 cm

area of triangle Q P R=$30cm{2}$

and P S is the new height for the triangle Q P R

For length P S

$area=\frac{1}{2}foot*height \\30=\frac{1}{2} *13*PS\\PS=\frac{30*2}{13} \\PS=\frac{60}{13} cm$

HENCE, the area of the triangle Q P R is $30cm^{2}$ and length P S is $\frac{60}{13} cm$