Question

in PQR, PR = 8cm, QR =4cm and PL = 5cm find (a) area of triangle PQR and (b) QM

Answer 1

find the value of LR
PL²+LR²=PR²
5² +LR² =8²
LR =√64-25
LR. =√39
LR=6.24cm
now,
LQ=6.24-4=2.24cm
in triangle PLQ

PL²+LQ²=PQ²
5²+2.24²=PQ²
25+5.01=PQ²
PQ=5.47cm

Area of triangle( PLR)=1/2×B×H
=1/2×6.24×5
=15.6 cm²

Area of triangle(PLQ)=1/2×B×H
=1/2×2.24×5
=5.6 cm²

area of ∆PQR=15.6-5.6
=10 cm²

1/2×8×QM=10
QM=10/4
QM=2.5 cm

Answer 2

Answer:

Area of triangle is 10 sq.cm and Height QM be 2.5 cm.

Step-by-step explanation:

GIVEN:

Refer figure:

$(i) \mathrm{QR}= base =4 \mathrm{~cm}, \mathrm{PL}= height =5 \mathrm{~cm} ,\mathram{PR}= 8cm$

Take QR as base and PL as height.

Area of the triangle

• $P Q R=\frac{1}{2} b h$

                $\begin{aligned}&=\frac{1}{2} \times 4 \times 5 \\&=10 \mathrm{~cm}^{2}\end{aligned}$

Case ii

$\mathrm{PR}= base =8 \mathrm{~cm}, \\\mathrm{QM}= height =? , \\Area =10 \mathrm{~cm}^{2} \\Area of triangle =\frac{1}{2} \times b \times h$

                     $&10=\frac{1}{2} \times 8 \times h \\\\&\mathrm{~h}=\frac{10}{4}=\frac{5}{2}=2.5 . \\\\&\mathrm{QM}=2.5 \mathrm{~cm}\end{aligned}$