Model Test Paper – Solved ഉത്തരങ്ങൾ

Draw AB = 7 cm. From A draw a ray at any convenient angle and step off three equal lengths on it with the compass (A₁, A₂, A₃). Join A₃ to B. Through A₂ and A₁ draw lines parallel to A₃B with set squares; they cut AB at Q and P. Four parallel lines (through A, A₁, A₂, A₃) cut the ray into equal parts, so they cut AB into equal parts too: AP = PQ = QB = 2⅓ cm.

Draw AB = 12 cm and a ray from A. Mark four equal steps on the ray: A₁ after one step and A₄ after four steps (1 + 3). Join A₄ to B and draw the line through A₁ parallel to A₄B; it meets AB at P. The parallels cut AB in the same ratio as the ray: AP : PB = 1 : 3, so AP = 12 × ¼ = 3 cm and PB = 12 × ¾ = 9 cm.

A line parallel to one side cuts the other two sides into the same fractions. AP is 2/8 = ¼ of AB, so AQ is ¼ of AC: AQ = ¼ × 6 = 1.5 cm, and QC = 6 − 1.5 = 4.5 cm. Check with the ratio form: AP : PB = 2 : 6 = 1 : 3 and AQ : QC = 1.5 : 4.5 = 1 : 3 ✔.

Each side of the midpoint quadrilateral joins the midpoints of two sides of a triangle whose third side is a diagonal, so it is half that diagonal (midpoint theorem). Two sides are ½ × 10 = 5 cm and two are ½ × 6 = 3 cm. Perimeter = 5 + 3 + 5 + 3 = 16 cm.

(i) The centroid divides the median 2 : 1 from the vertex: AG = ⅔ × 12 = 8 cm, GD = ⅓ × 12 = 4 cm.
(ii) The circumcentre of a right triangle is the midpoint of the hypotenuse, so its distance from each vertex is half the hypotenuse: 5 cm.

Let D be the midpoint of AB in triangle ABC, and let the line through D parallel to BC meet AC at E. Draw the line through A parallel to BC as well. Now three parallel lines (through A, through D, and BC) cut the two lines AB and AC, so they cut them in the same ratio: AD/DB = AE/EC. Since AD = DB, the ratio is 1, so AE = EC: the line bisects AC. ∎

(i) Draw the line through O parallel to AB (it is also parallel to DC). Then AB, this line and DC are three parallel lines cutting the two lines AC and BD. Three parallels cut any two lines in the same ratio, so AO/OC = BO/OD. ∎
(ii) BO/OD = AO/OC = 6/4 = 3/2, so BO and OD are in the ratio 3 : 2 and add up to 15 cm: BO = 15 × 3/5 = 9 cm, OD = 15 × 2/5 = 6 cm.

(i) Draw AB = 12 cm and a ray from A. Choose any unit and mark 3 units, then 4 units, then 5 units along the ray (points X, Y, Z). Join Z to B and draw lines through X and Y parallel to ZB; they cut AB at P and Q. The parallels through A, X, Y, Z are four parallel lines; parallel lines cut any two lines in the same ratio, so AP : PQ : QB = AX : XY : YZ = 3 : 4 : 5.
(ii) The sides are 12 × 3/12 = 3 cm, 12 × 4/12 = 4 cm and 12 × 5/12 = 5 cm. Taking PQ = 4 cm as the base, arcs of 3 cm from P and 5 cm from Q meet at the third vertex R. Since 3² + 4² = 9 + 16 = 25 = 5², it is a right triangle (right angle between the 3 cm and 4 cm sides).

(i) Let the right angle be at C, with CA = 6 cm and CB = 8 cm; M, N, L are the midpoints of CA, CB, AB. ML joins the midpoints of CA and AB, so ML ∥ CB and ML = ½CB = 4 cm. NL joins the midpoints of CB and AB, so NL ∥ CA and NL = ½CA = 3 cm. Both pairs of opposite sides of CMLN are parallel, so it is a parallelogram; its angle at C is 90°, so it is a rectangle.
(ii) Area of the rectangle = 3 × 4 = 12 sq cm. Area of the triangle = ½ × 6 × 8 = 24 sq cm. The rectangle is half the triangle.

In triangle ABC let D and E be the midpoints of BC and CA, and let the medians AD and BE meet at G. By the midpoint theorem, ED ∥ AB and ED = ½AB. Let P and Q be the midpoints of AG and BG; in triangle ABG the midpoint theorem gives PQ ∥ AB and PQ = ½AB. So PQ and ED are equal and parallel: PQDE is a parallelogram, and its diagonals PD and QE bisect each other: PG = GD and QG = GE. But AP = PG (P is the midpoint of AG), so AG = AP + PG = 2GD; likewise BG = 2GE. Thus the point where two medians meet divides each of them in the ratio 2 : 1 from the vertex.

Now take the medians BE and CF (F the midpoint of AB). By the same argument their meeting point divides BE in the ratio 2 : 1 from B – but there is only one such point on BE, namely G. So CF also passes through G: all three medians pass through G (the centroid, മധ്യമകേന്ദ്രം), and G divides each of them 2 : 1. ∎

(i) In quadrilateral ABCD let P, Q, R, S be the midpoints of AB, BC, CD, DA. Draw the diagonal AC. In triangle ABC, PQ joins the midpoints of two sides, so PQ ∥ AC and PQ = ½AC. In triangle ADC, SR ∥ AC and SR = ½AC. So PQ and SR are equal and parallel, and PQRS is a parallelogram. ∎ (Using the diagonal BD in the same way, QR and SP are each ½BD.)
(ii) The sides of PQRS are ½AC and ½BD. If the diagonals are equal, all four sides are equal – a rhombus. If the diagonals are perpendicular, then PQ (∥ AC) is perpendicular to QR (∥ BD), so the angles are right angles – a rectangle. Equal AND perpendicular diagonals give both: a square.
(iii) Each side of the midpoint quadrilateral is ½ × 8 = 4 cm and the angles are right angles, so it is a square of side 4 cm: area = 4 × 4 = 16 sq cm.