SSC (English Medium) इयत्ता १० वी - Maharashtra State Board Important Questions

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

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

Find QP using given information in the figure.

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

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

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

`"A(∆ ABD)"/"A(∆ ADC)"`

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

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. 

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"`.

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)]

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.) 

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

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

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)

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)]

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

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

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.

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

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

In the given, seg BE ⊥ seg AB and seg BA ⊥ seg AD.

if BE = 6 and AD = 9 find `(A(Δ ABE))/(A(Δ BAD))`.

In the figure, ray YM is the bisector of ∠XYZ, where seg XY ≅ seg YZ, find the relation between XM and MZ.