Advertisements
Advertisements
प्रश्न
In the given figure ∠BAP = ∠DCP = 70°, PC = 6 cm and CA = 4 cm, then PD : DB is ______.
विकल्प
5 : 3
3 : 5
3 : 2
2 : 3
MCQ
रिक्त स्थान भरें
उत्तर
In the given figure ∠BAP = ∠DCP = 70°, PC = 6 cm and CA = 4 cm, then PD : DB is 3 : 2.
Explanation:
In ΔAPB and ΔCPD,
∠A = ∠C = 70° ...(Given)
And ∠P = ∠P ...(Common)
∴ ΔAPB ∼ ΔCPD ...(By AA – axiom)
So, `(PD)/(DB) = (PC)/(CA) = 6/4 = 3/2`
∴ PD : DB = 3 : 2
shaalaa.com
Conditions for Similarity of Two Triangles: (Sas, Aa Or Aaa and Sss)
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
In the given diagram, ΔABC ∼ ΔPQR. If AD and PS are bisectors of ∠BAC and ∠QPR respectively then ______.
In the given diagram ΔADB and ΔACB are two right angled triangles with ∠ADB = ∠BCA = 90°. If AB = 10 cm, AD = 6 cm, BC = 2.4 cm and DP = 4.5 cm.
- Prove that ΔAPD ∼ ΔBPC
- Find the length of BD and PB
- Hence, find the length of PA
- Find area ΔAPD : area ΔBPC.