Advertisements
Advertisements
प्रश्न
In the given figure, ∠B = ∠E, ∠ACD = ∠BCE, AB = 10.4 cm and DE = 7.8 cm. Find the ratio between areas of the ∆ABC and ∆DEC.
उत्तर
Given, ∠ACD = ∠BCE
∠ACD + ∠BCD = ∠BCE + ∠BCD
∠ACB = ∠DCE
Also, given ∠B = ∠E
∴ ∆ABC ∼ ∆DEC
`(ar(ΔABC))/(ar(ΔDEC)) = ((AB)/(DE))^2`
= `(10.4/7.8)^2`
= `(4/3)^2`
= `16/9`
APPEARS IN
संबंधित प्रश्न
Given: RS and PT are altitudes of ΔPQR. Prove that:
- ΔPQT ~ ΔQRS,
- PQ × QS = RQ × QT.
In ∆ABC, ∠B = 90° and BD ⊥ AC.
- If CD = 10 cm and BD = 8 cm; find AD.
- If AC = 18 cm and AD = 6 cm; find BD.
- If AC = 9 cm and AB = 7 cm; find AD.
In the right-angled triangle QPR, PM is an altitude.
Given that QR = 8 cm and MQ = 3.5 cm, calculate the value of PR.
In the given figure, AX : XB = 3 : 5
Find:
- the length of BC, if the length of XY is 18 cm.
- the ratio between the areas of trapezium XBCY and triangle ABC.
In the given triangle PQR, LM is parallel to QR and PM : MR = 3 : 4.
Calculate the value of ratio:
- `(PL)/(PQ)` and then `(LM)/(QR)`
- `"Area of ΔLMN"/"Area of ΔMNR"`
- `"Area of ΔLQM"/"Area of ΔLQN"`
Triangle ABC is an isosceles triangle in which AB = AC = 13 cm and BC = 10 cm. AD is
perpendicular to BC. If CE = 8 cm and EF ⊥ AB, find:
i)`"area of ADC"/"area of FEB"` ii)`"area of ΔAFEB"/"area of ΔABC"`
In the given figure, triangle ABC is similar to triangle PQR. AM and PN are altitudes whereas AX and PY are medians.
Prove that : `(AM)/(PN)=(AX)/(PY)`
In a triangle PQR, L and M are two points on the base QR, such that ∠LPQ = ∠QRP and ∠RPM = ∠RQP. Prove that:
- ΔPQL ∼ ΔRPM
- QL × RM = PL × PM
- PQ2 = QR × QL
In the give figure, ABC is a triangle with ∠EDB = ∠ACB. Prove that ΔABC ∼ ΔEBD. If BE = 6 cm, EC = 4 cm, BD = 5 cm and area of ΔBED = 9 cm2. Calculate the:
- length of AB
- area of ΔABC
In the given figure, ABC is a right angled triangle with ∠BAC = 90°.
- Prove that : ΔADB ∼ ΔCDA.
- If BD = 18 cm and CD = 8 cm, find AD.
- Find the ratio of the area of ΔADB is to area of ΔCDA.
