हिंदी

In the Given Figure, ∠1 = ∠2 and `(Ac)/(Bd)=(Cb)/(Ce)` Prove that δ Acb ~ δ Dce. - Mathematics

Advertisements
Advertisements

प्रश्न

In the given figure, ∠1 = ∠2 and `(AC)/(BD)=(CB)/(CE)`   Prove that Δ ACB ~ Δ DCE. 

उत्तर

We have : 

`(AC)/(BD)=(CB)/(CE)` 

⟹ `(AC)/(CB)=(CD)/(CE)` (𝑆𝑖𝑛𝑐𝑒,𝐵𝐷=𝐷𝐶 𝑎𝑠 ∠1= ∠2 )
Also, ∠1= ∠2
i.e, ∠𝐷𝐵𝐶=∠𝐴𝐶𝐵
Therefore, by SAS similarity theorem, we get :
Δ ACB - Δ DCE 

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Triangles - Exercises 2

APPEARS IN

आरएस अग्रवाल Mathematics [English] Class 10
अध्याय 4 Triangles
Exercises 2 | Q 15

संबंधित प्रश्न

In the following figure, in Δ PQR, seg RS is the bisector of ∠PRQ.

PS = 3, SQ = 9, PR = 18. Find QR.


In figure, ∠A = ∠CED, prove that ∆CAB ~ ∆CED. Also, find the value of x.


In the following figure, seg DH ⊥ seg EF and seg GK ⊥ seg EF. If DH = 18 cm, GK = 30 cm and `A(triangle DEF) = 450 cm^2`, then find:

1) EF

2) `A(triangle GFE)`

3) `A(square DFGE)`


Prove that, in a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.


State, true or false:

Two similar polygons are necessarily congruent.


State, true or false:

All isosceles triangles are similar.


In the given figure, BC is parallel to DE. Area of triangle ABC = 25 cm2, Area of trapezium BCED = 24 cm2 and DE = 14 cm. Calculate the length of BC. Also, find the area of triangle BCD.


The areas of two similar triangles are `64cm^2` and `100cm^2` respectively. If a median of the smaller triangle is 5.6cm, find the corresponding median of the other. 


In ∆ABC, AP ⊥ BC, BQ ⊥ AC B– P–C, A–Q – C then prove that, ∆CPA ~ ∆CQB. If AP = 7, BQ = 8, BC = 12 then Find AC. 


In figure , DEF is a right -angled triangle with ∠ E = 90 °.FE is produced to G and GH is drawn perpendicular to DE = 8 cm , DH = 8 cm ,DH = 6 cm and HF = 4 cm , find `("Ar" triangle "DEF")/("Ar" triangle "GHF")`


The dimensions of a buiIding are 50 m Iong, 40m wide and 70m high. A model of the same building is made with a scale factor of 1: 500. Find the dimensions of the model.


The scale of a map is 1 : 200000. A plot of land of area 20km2 is to be represented on the map. Find

The area on the map that represents the plot of land. 


The actual area of an island is 1872km2 . On a map, this area is 117 cm2. if the length of the coastline is 44cm on the map, find the length of its actual coastline. 


In the given figure, PQ || AB; CQ = 4.8 cm QB = 3.6 cm and AB = 6.3 cm. Find :

  1. `(CP)/(PA)` 
  2. PQ
  3. If AP = x, then the value of AC in terms of x.


In ΔABC, point D divides AB in the ratio 5:7, Find: BC, If DE = 2.5cm


In a quadrilateral PQRS, the diagonals PR and QS intersect each other at the point T. If PT:TR = QT :TS = 1:2, show that ΔPTQ - DRTS


Harmeet is 6 feet tall and casts a shadow of 3 feet long. What is the height of a nearby pole if it casts a shadow of 12 feet long at the same time?


In ΔABC, DE is drawn parallel to BC cutting AB in the ratio 2 : 3. Calculate:
(i) `("area"(Δ"ADE"))/("area"(Δ"ABC")`

(i) `("area"("trapeziumEDBC"))/("area"(Δ"ABC"))`


The dimensions of the model of a building are 1.2m x 75cm x 2m. If the scale factor is 1 : 20; find the actual dimensions of the building.


A model of cargo tuck is made to a scale of 1:40. The length of the model is 15cm. Calculate: The volume of the model if the volume of the truck is 6m3


Check whether the triangles are similar and find the value of x


In the adjacent figure, ∆ABC is right angled at C and DE ⊥ AB. Prove that ∆ABC ~ ∆ADE and hence find the lengths of AE and DE


If ΔABC ~ ΔLMN and ∠B = 40°, then ∠M = ? Give reason.


In triangle ABC point D is on side BC (B−D−C) such that ∠BAC = ∠ADC then prove that CA2 = CB × CD


In a square of side 10 cm, its diagonal = ______.


∆ABC ~ ∆PQR. If AM and PN are altitudes of ΔABC and ∆PQR respectively and AB2 : PQ2 = 4 : 9, then AM : PN = ______.


If ∆ABC ~ ∆QRP, `(ar(ABC))/(ar(PQR)) = 9/4`, AB = 18 cm and BC = 15 cm, then PR is equal to ______.


Prove that if a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.

Using the above theorem prove that a line through the point of intersection of the diagonals and parallel to the base of the trapezium divides the non-parallel sides in the same ratio.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×