SSC (Marathi Semi-English) इयत्ता १० वी - Maharashtra State Board Question Bank Solutions for Geometry Mathematics 2

Given below is the triangle and length of line segments. Identify in the given figure, ray PM is the bisector of ∠QPR.

Given below is the triangle and length of line segments. Identify in the given figure, ray PM is the bisector of ∠QPR.

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

Measures of some angles in the figure are given. Prove that `"AP"/"PB" = "AQ"/"QC"`.

Find QP using given information in the figure.

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

In ∆LMN, ray MT bisects ∠LMN If LM = 6, MN = 10, TN = 8, then Find LT. 

In ∆ABC, seg BD bisects ∠ABC. If AB = x, BC = x + 5, AD = x – 2, DC = x + 2, then find the value of x.

∆PQR ~ ∆LTR. In ∆PQR, PQ = 4.2 cm, QR = 5.4 cm, PR = 4.8 cm. Construct ∆PQR and ∆LTR, such that `"PQ"/"LT" = 3/4`.

∆AMT ~ ∆AHE. In ∆AMT, AM = 6.3 cm, ∠TAM = 50°, AT = 5.6 cm. `"AM"/"AH" = 7/5`. Construct ∆AHE.

∆ABC ~ ∆LBN. In ∆ABC, AB = 5.1 cm, ∠B = 40°, BC = 4.8 cm, \[\frac{AC}{LN} = \frac{4}{7}\]. Construct ∆ABC and ∆LBN.

Construct ∆PYQ such that, PY = 6.3 cm, YQ = 7.2 cm, PQ = 5.8 cm. If \[\frac{YZ}{YQ} = \frac{6}{5},\] then construct ∆XYZ similar to ∆PYQ.

Find the ratio in which point P(k, 7) divides the segment joining A(8, 9) and B(1, 2). Also find k.

If A(–14, –10), B(6, –2) is given, find the coordinates of the points which divide segment AB into four equal parts.

If A (20, 10), B(0, 20) are given, find the coordinates of the points which divide segment AB into five congruent parts.

In ∆PQR seg PM is a median. Angle bisectors of ∠PMQ and ∠PMR intersect side PQ and side PR in points X and Y respectively. Prove that XY || QR. 

Complete the proof by filling in the boxes.

In △PMQ, ray MX is bisector of ∠PMQ.

∴ `square/square = square/square` .......... (I) theorem of angle bisector.

In △PMR, ray MY is bisector of ∠PMQ.

∴ `square/square = square/square` .......... (II) theorem of angle bisector.

But `(MP)/(MQ) = (MP)/(MR)` .......... M is the midpoint QR, hence MQ = MR.

∴ `(PX)/(XQ) = (PY)/(YR)`

∴ XY || QR .......... converse of basic proportionality theorem.

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 ▢ABCD, seg AD || seg BC. Diagonal AC and diagonal BD intersect each other in point P. Then show that `"AP"/"PD" = "PC"/"BP"`.