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Geometry Mathematics 2
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In the given figure, BC ⊥ AB, AD ⊥ AB, BC = 4, AD = 8, then find `("A"(∆"ABC"))/("A"(∆"ADB"))`

Appears in 1 question paper
Chapter: [1] Similarity
Concept: Properties of Ratios of Areas of Two Triangles

In adjoining figure, PQ ⊥ BC, AD ⊥ BC then find following ratios.

  1. `("A"(∆"PQB"))/("A"(∆"PBC"))`
  2. `("A"(∆"PBC"))/("A"(∆"ABC"))`
  3. `("A"(∆"ABC"))/("A"(∆"ADC"))`
  4. `("A"(∆"ADC"))/("A"(∆"PQC"))`
Appears in 1 question paper
Chapter: [1] Similarity
Concept: Properties of Ratios of Areas of Two Triangles

In ∆PQR, PM = 15, PQ = 25 PR = 20, NR = 8. State whether line NM is parallel to side RQ. Give reason. 

Appears in 1 question paper
Chapter: [1] Similarity
Concept: Property of Three Parallel Lines and Their Transversals

In ∆MNP, NQ is a bisector of ∠N. If MN = 5, PN = 7 MQ = 2.5 then find QP. 

Appears in 1 question paper
Chapter: [1] Similarity
Concept: Property of an Angle Bisector of a Triangle

In trapezium ABCD, side AB || side PQ || side DC, AP = 15, PD = 12, QC = 14, Find BQ. 

Appears in 1 question paper
Chapter: [1] Similarity
Concept: Property of Three Parallel Lines and Their Transversals

In the given figure, if AB || CD || FE then find x and AE. 

Appears in 1 question paper
Chapter: [1] Similarity
Concept: Property of an Angle Bisector of a Triangle

 In trapezium PQRS, side PQ || side SR, AR = 5AP, AS = 5AQ then prove that, SR = 5PQ 

 

 

Appears in 1 question paper
Chapter: [1] Similarity
Concept: Properties of Ratios of Areas of Two Triangles

In ∆ABC, B - D - C and BD = 7, BC = 20 then find following ratio. 

`"A(∆ ABD)"/"A(∆ ADC)"`

Appears in 1 question paper
Chapter: [1] Similarity
Concept: Properties of Ratios of Areas of Two Triangles

In the given figure, ∠ABC = ∠DCB = 90° AB = 6, DC = 8 then `(A(Δ ABC))/(A(Δ DCB))` = ?

Appears in 1 question paper
Chapter: [1] Similarity
Concept: Properties of Ratios of Areas of Two Triangles

In the given figure, seg PA, seg QB, seg RC, and seg SD are perpendicular to line AD.

AB = 60, BC = 70, CD = 80, PS = 280 then find PQ, QR, and RS. 

Appears in 1 question paper
Chapter: [1] Similarity
Concept: Property of Three Parallel Lines and Their Transversals

In the given fig, bisectors of ∠B and ∠C of ∆ABC intersect each other in point X. Line AX intersects side BC in point Y. AB = 5, AC = 4, BC = 6 then find `"AX"/"XY"`.

Appears in 1 question paper
Chapter: [1] Similarity
Concept: Property of an Angle Bisector of a Triangle

In the given fig, XY || seg AC. If 2AX = 3BX and XY = 9. Complete the activity to Find the value of AC.  

Activity: 2AX = 3BX  

∴ `"AX"/"BX" = square/square`

`("AX" +"BX")/"BX" = (square + square)/square`   ...(by componendo)

`"AB"/"BX" = square/square`           ...(I)

ΔBCA ~ ΔBYX             ...`square` test of similarity,

∴ `"BA"/"BX" = "AC"/"XY"`    ...(corresponding sides of similar triangles)

∴ `square/square = "AC"/9`      

∴ AC = `square`        ...[From(I)]

Appears in 1 question paper
Chapter: [1] Similarity
Concept: Property of Three Parallel Lines and Their Transversals

In the given figure, the vertices of square DEFG are on the sides of ∆ABC. ∠A = 90°. Then prove that DE2 = BD × EC. (Hint: Show that ∆GBD is similar to ∆CFE. Use GD = FE = DE.) 

Appears in 1 question paper
Chapter: [1] Similarity
Concept: Property of Three Parallel Lines and Their Transversals

In Δ ABC and Δ PQR,
∠ ABC ≅ ∠ PQR, seg BD and
seg QS are angle bisector.
`If  (l(AD))/(l(PS)) = (l(DC))/(l(SR))`
Prove that : Δ ABC ∼ Δ PQR

Appears in 1 question paper
Chapter: [1] Similarity
Concept: Property of an Angle Bisector of a Triangle

Seg NQ is the bisector of ∠ N
of Δ MNP. If MN= 5, PN =7,
MQ = 2.5 then find QP.

Appears in 1 question paper
Chapter: [1] Similarity
Concept: Property of an Angle Bisector of a Triangle

From the top of a light house, an abserver looking at a boat makes an angle of depression of 600. If the height of the lighthouse is 90 m then find how far is the boat from the lighthouse. (3 = 1.73)

Appears in 1 question paper
Chapter: [1] Similarity
Concept: Property of an Angle Bisector of a Triangle

In ΔABC, ray BD bisects ∠ABC.

If A – D – C, A – E – B and seg ED || side BC, then prove that:

`("AB")/("BC") = ("AE")/("EB")`

Proof : 

In ΔABC, ray BD bisects ∠ABC.

∴ `("AB")/("BC") = (......)/(......)`   ......(i) (By angle bisector theorem)

In ΔABC, seg DE || side BC

∴ `("AE")/("EB") = ("AD")/("DC")`   ....(ii) `square`

∴ `("AB")/square = square/("EB")`   [from (i) and (ii)]

Appears in 1 question paper
Chapter: [1] Similarity
Concept: Property of an Angle Bisector of a Triangle

In ΔABC, ∠ACB = 90°. seg CD ⊥ side AB and seg CE is angle bisector of ∠ACB.

Prove that: `(AD)/(BD) = (AE^2)/(BE^2)`.

Appears in 1 question paper
Chapter: [1] Similarity
Concept: Property of an Angle Bisector of a Triangle

The ratio of the areas of two triangles with the common base is 4 : 3. Height of the larger triangle is 2 cm, then find the corresponding height of the smaller triangle.

Appears in 1 question paper
Chapter: [1] Similarity
Concept: Properties of Ratios of Areas of Two Triangles

In ∆ABC, B – D – C and BD = 7, BC = 20, then find the following ratio.

`(A(triangleABD))/(A(triangleABC))`

Appears in 1 question paper
Chapter: [1] Similarity
Concept: Properties of Ratios of Areas of Two Triangles
< prev  41 to 60 of 329  next > 
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