Question

find the area of triangle LOK whereLO=LK = 5cm and LB=3cm
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Answer 1

Answer:

Area of ΔLOK = 12cm²

Step-by-step explanation:

LO = LK = 5cm

LB = 3cm

The perpendicular height of an isosceles triangle is a perpendicular bisector of its base.

OB = BK = $\sqrt{LK^{2} - LB^{2} }$

              = $\sqrt{5^{2} - 3^{2} }$

              = $\sqrt{25-9}$

              = $\sqrt{16}$

              = 4cm

OK = OB + BK = 4+4 = 8cm

Area =$\frac{1}{2}$bh = $\frac{1}{2}$ × 8 × 3 = 12cm²//

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Answer 2

Given :-

Here , The side LO and LK equal to 5 cm

The height LB = 3cm

Solution :-

Let take the triangle LBO

Here , LO = 5cm , LB = 3cm , OB = ?

By Using phythagores theorem ,

H^2 = P^2 + B^2

( LO)^2 = ( LB)^2 + ( BO) ^2

( 5 )^2 = ( 3 )^2 + ( BO )^2

25 = 9 + ( B0 )^2

( B0 )^2 = 25 - 9

( B0 )^2 = 16cm

B0 = 4cm

Let take the triangle LBK

Here, LK = 5cm , LB = 3cm and BK = ?

By Using phythagores theorem ,

H^2 = P^2 + B^2

( LK )^2 = ( LB )^2 + ( BK )^2

( 5 )^2 = ( 3 )^2 + ( BK )^2

25 = 9 + ( BK )^2

( BK )^2 = 25 - 9

( BK )^2 = 16

BK = 4cm

We know that , OK = BO + BK

Therefore , OK = 5 + 5 = 10 cm

Area of triangle

= 1/2 * Base * Height

Put the required values in the formula ,

Area of triangle = 1/2 * 10 * 3

Area of triangle = 5 * 3

Area of triangle = 15 cm^2 $.$