In trapezium ABCD, AB∥DC. The sides AD and BC are produced to meet at E. AB = 15 cm, CD = 10 cm, AD = 4 cm and BC = 5 cm, then DE = ?
8 cm
The correct answer is 8 cm.
Express other relevant lengths in terms of this variable (e.g., EA = ED + DA = x + 4). Form an equation using the proportional ratios involving the known lengths and the variable. Solve the algebraic equation for the variable. This method is widely applicable to geometry problems involving similar figures formed by parallel lines and transversals.
It is given that ΔABC ~ ΔXYZ and Area ΔABC : Area ΔXYZ = 81 : 25. If AB = 18 cm, BC = 10 cm, CA = 15 cm, then what is the side XZ (in cm)?
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Let ΔABC ~ ΔPQR and \(\frac{{ar\left( {{\rm{\Delta \;ABC}}} \right)}}{{ar\left( {{\rm{\Delta \;PQR}}} \right)}} = \frac{9}{{169}}\) . If AB = 12 cm, BC = 7 cm and AC = 8 cm, then PQ (in cm) is equal to:
The side BC Δ ABC is produced to D. The bisectors of ∠ABC and ∠ACD meet at E. If AB = AC and ∠BEC = 24º, then the measure of ∠ABC is:
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It is given that ΔABC ~ ΔXYZ and Area ΔABC : Area ΔXYZ = 81 : 25. If AB = 18 cm, BC = 10 cm, CA = 15 cm, then what is the side XZ (in cm)?
Sides of two similar triangles are in the ratio 4 ∶ 9. Area of these triangles are in the ratio:
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