Question

bcdef ~ kjihg. what is ( mangle d )? ( mangle d = square^circ )

Explanation:

Step1: Recall sum of interior angles of a hexagon? Wait, no, B C D E F and K J I H G – wait, the figures: B C D E F – let's count the sides. Wait, B C D E F – that's 5 sides? Wait, no, the first figure: points B, C, D, E, F – so pentagon? Wait, B C D E F: 5 vertices, so pentagon. Similarly, K J I H G: 5 vertices, pentagon. So both are pentagons. Sum of interior angles of a pentagon is (5 - 2)*180° = 540°.

Step2: Find corresponding angles. Since B C D E F ~ K J I H G, the order of the letters matters. So B corresponds to J, C to I? Wait, no, the similarity is B C D E F ~ K J I H G. So B ↔ K? Wait, no, the notation: B C D E F ~ K J I H G, so the order is B→K, C→J, D→I, E→H, F→G? Wait, no, let's check the angles. In the second pentagon (K J I H G), angles are at K: 132°, J: 138°, I: 75°, H:?, G:? Wait, no, the first pentagon: B C D E F, angles at C: 138°, F: 91°, and we need angle D. The second pentagon: K J I H G, angles at K: 132°, J: 138°, I: 75°, and we can find the other angles.

Wait, sum of interior angles of pentagon is 540°. Let's calculate the sum of known angles in the second pentagon (K J I H G): 132° (K) + 138° (J) + 75° (I) +? (H) +? (G)? Wait, no, maybe the first pentagon: B C D E F, angles at C: 138°, F: 91°, and we need angle D. The second pentagon: K J I H G, angles at K: 132°, J: 138°, I: 75°, and let's find the missing angles. Wait, maybe the correspondence is B ↔ J, C ↔ I, D ↔ H, E ↔ K, F ↔ G? Wait, no, let's list the angles.

Wait, first pentagon (B C D E F): angles at C: 138°, F: 91°, and we need angle D. Second pentagon (K J I H G): angles at K: 132°, J: 138°, I: 75°, and let's compute the sum. Sum is 540°. So in the second pentagon, sum of known angles: 132 (K) + 138 (J) + 75 (I) +? (H) +? (G)? Wait, no, maybe the first pentagon's angles: let's denote the angles as ∠B, ∠C, ∠D, ∠E, ∠F. Second pentagon: ∠K, ∠J, ∠I, ∠H, ∠G. Since they are similar, corresponding angles are equal. So let's find the correspondence.

Looking at the angles: ∠C in first pentagon is 138°, ∠J in second is 138°, so C ↔ J. ∠F in first is 91°, maybe ∠G in second? Wait, no, let's calculate the sum for the second pentagon. Let's assume the second pentagon has angles: K:132, J:138, I:75, H:?, G:?. Sum is 132 + 138 + 75 + H + G = 540. So 132+138=270, 270+75=345. So H + G = 540 - 345 = 195. Now, the first pentagon: angles at C:138 (corresponds to J:138), F:91, and we need angle D. Let's see the first pentagon's angles: ∠B, ∠C=138, ∠D, ∠E, ∠F=91. Sum is 540. So ∠B + 138 + ∠D + ∠E + 91 = 540 → ∠B + ∠D + ∠E = 540 - 138 - 91 = 311.

Now, in the second pentagon, ∠K=132, ∠J=138, ∠I=75, so ∠H and ∠G. Wait, maybe the correspondence is B ↔ K, C ↔ J, D ↔ I, E ↔ H, F ↔ G. So ∠B corresponds to ∠K=132°, ∠C=138° corresponds to ∠J=138°, ∠D corresponds to ∠I=75°? No, that can't be, because in the first pentagon, angle at F is 91°, maybe ∠F corresponds to ∠G. Wait, let's check the sum again.

Wait, maybe I made a mistake. Let's re-express:

First pentagon (B C D E F): vertices B, C, D, E, F (order matters for similarity).

Second pentagon (K J I H G): vertices K, J, I, H, G.

So similarity: B ↔ K, C ↔ J, D ↔ I, E ↔ H, F ↔ G.

Therefore, ∠B = ∠K = 132°, ∠C = ∠J = 138°, ∠D = ∠I = 75°? Wait, no, ∠I is 75°, but in the first pentagon, angle at F is 91°, so ∠F = ∠G. Let's calculate sum for first pentagon:

∠B (K's angle:132°) + ∠C (138°) + ∠D (?) + ∠E (H's angle) + ∠F (91°) = 540°.

Sum of known angles in first pentagon: 132 + 138 + 91 = 361. So ∠D + ∠E = 540 - 361 = 179.

In the second pentagon: ∠K=132, ∠J=138, ∠I=75, ∠H=?, ∠G=91 (since ∠F=91 and F↔G). So sum of second pentagon: 132 + 138 + 75 + ∠H + 91 = 540. 132+138=270, 270+75=345, 345+91=436. So ∠H = 540 - 436 = 104? No, that doesn't match. Wait, maybe the correspondence is different.

Wait, maybe the first pentagon is B C D E F, and the second is K J I H G, so the order is B→J, C→I, D→H, E→G, F→K? No, this is confusing. Wait, let's count the angles in the second pentagon. The second pentagon has angles at K:132°, J:138°, I:75°, and two other angles. Wait, maybe the first pentagon: B C D E F, angles at C:138°, F:91°, and we need angle D. The second pentagon: K J I H G, angles at K:132°, J:138°, I:75°, and let's find ∠H and ∠G.

Sum of pentagon angles: 540. So in second pentagon: 132 + 138 + 75 + ∠H + ∠G = 540 → 345 + ∠H + ∠G = 540 → ∠H + ∠G = 195.

In first pentagon: ∠B + 138 (∠C) + ∠D + ∠E + 91 (∠F) = 540. If ∠B corresponds to ∠H, ∠E corresponds to ∠G, then ∠B + ∠E = 195 (since ∠H + ∠G=195). And ∠D corresponds to ∠I=75°? Wait, ∠I is 75°, so ∠D=75°? But let's check again.

Wait, maybe the first pentagon's angles: ∠B, ∠C=138°, ∠D=?, ∠E=?, ∠F=91°. The second pentagon's angles: ∠K=132°, ∠J=138°, ∠I=75°, ∠H=?, ∠G=?. Since they are similar, the angles must match in order. So B C D E F ~ K J I H G, so B→K, C→J, D→I, E→H, F→G. Therefore:

∠B = ∠K = 132°,

∠C = ∠J = 138°,

∠D = ∠I = 75°,

∠E = ∠H,

∠F = ∠G.

Now, sum of first pentagon: ∠B (132) + ∠C (138) + ∠D (75) + ∠E + ∠F (91) = 540.

Calculate sum: 132 + 138 = 270; 270 + 75 = 345; 345 + 91 = 436. Then ∠E = 540 - 436 = 104. Then ∠H (which is ∠E) would be 104, and ∠G (which is ∠F) is 91. Now check second pentagon sum: ∠K (132) + ∠J (138) + ∠I (75) + ∠H (104) + ∠G (91) = 132+138=270; 270+75=345; 345+104=449; 449+91=540. Perfect, that works. So ∠D corresponds to ∠I, which is 75°? Wait, no, ∠I is 75°, so ∠D=75°? Wait, but let's confirm the correspondence again. The similarity is B C D E F ~ K J I H G, so the order is B to K, C to J, D to I, E to H, F to G. So angle D is at position D, which corresponds to angle I at position I. Therefore, ∠D = ∠I = 75°? Wait, but in the first pentagon, angle at F is 91°, which corresponds to angle G in the second pentagon, and angle at B corresponds to angle K (132°), angle at C to J (138°), angle D to I (75°), angle E to H, angle F to G. Then sum checks out. So m∠D = 75°? Wait, but let's check again.

Wait, maybe I mixed up the correspondence. Let's list the angles:

First pentagon (B C D E F):

- ∠B: corresponds to ∠K (132°)

- ∠C: corresponds to ∠J (138°)

- ∠D: corresponds to ∠I (75°)

- ∠E: corresponds to ∠H

- ∠F: corresponds to ∠G (91°)

Sum of first pentagon: 132 + 138 + 75 + ∠E + 91 = 540.

132 + 138 = 270; 270 + 75 = 345; 345 + 91 = 436. So ∠E = 540 - 436 = 104. Then ∠H (which is ∠E) is 104, and ∠G (which is ∠F) is 91. Then sum of second pentagon: 132 (K) + 138 (J) + 75 (I) + 104 (H) + 91 (G) = 132+138=270; 270+75=345; 345+104=449; 449+91=540. Perfect, that's correct. So ∠D corresponds to ∠I, which is 75°, so m∠D = 75°? Wait, but let's check the first pentagon's angle at D. Wait, maybe I made a mistake in correspondence. Wait, the first pentagon: B C D E F – the order is B to C to D to E to F to B. The second pentagon: K J I H G – K to J to I to H to G to K. So the order is B→K, C→J, D→I, E→H, F→G. So the angle at D is between C and E, and angle at I is between J and H. So yes, ∠D corresponds to ∠I, which is 75°. Therefore, m∠D = 75°? Wait, but let's check again. Wait, in the first pentagon, angle at C is 138°, which matches angle at J (138°), angle at B (which is ∠B) matches angle at K (132°), angle at F (91°) matches angle at G, angle at D matches angle at I (75°), and angle at E matches angle at H (104°). So that works. So m∠D is 75°? Wait, but let's confirm the sum again. 132 (B) + 138 (C) + 75 (D) + 104 (E) + 91 (F) = 132+138=270, 270+75=345, 345+104=449, 449+91=540. Correct. And the second pentagon: 132 (K) + 138 (J) + 75 (I) + 104 (H) + 91 (G) = 540. Correct. So ∠D is 75°? Wait, but the problem says "BCDEF ~ KJIHG". So B C D E F ~ K J I H G, so the order is B→K, C→J, D→I, E→H, F→G. Therefore, ∠D corresponds to ∠I, which is 75°, so m∠D = 75°.

Answer:

75