PQ is a line parallel to side BC and passing through vertex A of a triangle ABC. If BE $∥$ AC and CF $∥$ AB meet PQ at E and F respectively. If the base and altitude of $△$ ABC are 6 cm and 8 cm respectively, Find the area of $△$ ACF. [1 MARK]

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Solution

Concept : 0.5 Mark
Calculation : 0.5 Mark
CF when drawn parallel to AB , forms parallelogram ABCF.

AC is the diagonal of the parallelogram that divides ABCF into 2 equal triangles ABC and ACF.

So, Area of $△$ ACF = Area of $△$ ABC= $\frac{1}{2}$ × 6 × 8 = 24 $c m^{2}$

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PQ is a line parallel to side BC and passing through vertex A of a triangle ABC. If BE ∥ AC and CF ∥ AB meet PQ at E and F respectively. If the base and altitude of △ABC are 6 cm and 8 cm respectively, Find the area of △ACF. [1 MARK]

Concept : 0.5 Mark Calculation : 0.5 Mark CF when drawn parallel to AB , forms parallelogram ABCF. AC is the diagonal of the parallelogram that divides ABCF into 2 equal triangles ABC and ACF. So, Area of △ACF = Area of △ABC= 12 × 6 × 8 = 24 cm2

PQ is a line parallel to side BC and passing through vertex A of a triangle ABC. If BE $∥$ AC and CF $∥$ AB meet PQ at E and F respectively. If the base and altitude of $△$ ABC are 6 cm and 8 cm respectively, Find the area of $△$ ACF. [1 MARK]

Concept : 0.5 Mark
Calculation : 0.5 Mark
CF when drawn parallel to AB , forms parallelogram ABCF.

AC is the diagonal of the parallelogram that divides ABCF into 2 equal triangles ABC and ACF.

So, Area of $△$ ACF = Area of $△$ ABC= $\frac{1}{2}$ × 6 × 8 = 24 $c m^{2}$