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2) Suppose the line n intersect line l at K and line m at L. Since, ∠a ≅ ∠b, then m∠a = m∠b. It is known that, if a pair of alternate interior angles formed by a transversal of two lines is congruent, then the two lines are parallel. ∴ AB || CD or line l || line m.
3) Consider the longer of the two sides PQ and PR. Let's say that it is PQ (though it doesn't matter, and you can pick either if they are equal).
PS must be shorter than or equal to PQ. (It's only equal if S and Q are the same point.)
It is a rule that the sum of any two sides of a triangle must be greater than the third side. So we also know that PR + QR > PQ
Since PR + QR > PQ
We know PQ + PR + QR > 2PQ
(just added PQ to both sides)
And since PQ >= PS,
We know 2PQ >= 2PS
(multipled both sides by 2)
From those two (PQ + PR + QR > 2PQ, 2PQ >= 2PS), we know:
PQ + PR + QR > 2PS
4) Corresponding sides of the triangles:
XY : PQ, YZ : QR, ZX : RP.
For similar triangles ratios of corresponding sides are the same
XY : PQ = 4 : 8 = 1 : 2
YZ : QR = 1 : 2
ZX : RP = 1 : 2
YZ : QR = 1 : 2
6 : QR = 1 :2, QR = 12 (cm)
ZX : RP = 1 : 2
5 : RP = 1 : 2, RP = 10 (cm)
5) The measures of the angles of a triangle are x°, (x – 20)°, (x – 40)°. [Given] ∴ x°+ (x – 20)° + (x – 40)° = 180° [Sum of the measures of the angles of a triangle is 180°] ∴ 3x – 60 = 180 ∴ 3x = 180 + 60 ∴ 3x = 240 ∴ x = 240 ∴ x = 240/3 ∴ x = 80° ∴ The measures of the remaining angles are x – 20° = 80° – 20° = 60°, x – 40° = 80° – 40° = 40° ∴ The measures of the angles of the triangle are 80° , 60° and 40°.